Problem: In the diagram, $PQRS$ is a trapezoid with an area of $12.$ $RS$ is twice the length of $PQ.$ What is the area of $\triangle PQS?$

[asy]
draw((0,0)--(1,4)--(7,4)--(12,0)--cycle);
draw((7,4)--(0,0));
label("$S$",(0,0),W);
label("$P$",(1,4),NW);
label("$Q$",(7,4),NE);
label("$R$",(12,0),E);
[/asy]
Answer: Since $PQ$ is parallel to $SR,$ the height of $\triangle PQS$ (considering $PQ$ as the base) and the height of $\triangle SRQ$ (considering $SR$ as the base) are the same (that is, the vertical distance between $PQ$ and $SR$).

Since $SR$ is twice the length of $PQ$ and the heights are the same, the area of $\triangle SRQ$ is twice the area of $\triangle PQS.$ In other words, the area of $\triangle PQS$ is $\frac{1}{3}$ of the total area of the trapezoid, or $\frac{1}{3}\times 12 = \boxed{4}.$